12 NCERT MATH SUBJECTIVE QUESTIONS

Q1. Define conditional probability Solution Conditional Probability: If A and B are two events of sample space S associated with a random experiment, then occurrence of A provided B has already occurred is called conditional probability. It is denoted by P(A | B).   Q2. Define posteriori probability Solution Posteriori Probability: If E 1 , E 2 , ..., E n are the partitions of a sample space S, then the probability P(E i ) is called the priori probability of the hypothesis E i  and the conditional probability P(E i |A) is called a posteriori probability of the hypothesis E i .   Q3. Define Bernoulli trials. Solution Bernoulli Trials: Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes: success or failure. (iv) The probability of success remains the same in each trial ...

Q1. The foot of the perpendicular drawn from the origin to a plane is (2, 1, 5). Find the equation of the plane. Solution Since, the foot of the perpendicular to the plane is A(2, 1, 5). Therefore, (2, 1, 5) is the point on the plane. So, equation of the plane passing through the point (2,.1, 5) is: a(x – 2) + b(y – 1) + c(z – 5) = 0. Now, the direction ratios of the perpendicular line OA = 2 – 0, 1 – 0, 5 – 0, i.e., 2, 1, 5.   Therefore, the required plane is: 2(x – 2) + 1(y – 1) + 5(z – 5) = 0 i.e, 2x + y + 5z = 30.     Q2. Show that the planes 3x + 4y – 5z + 7 = 0 and x + 3y + 3z + 7 = 0 are perpendicular. Solution The direction ratios of the normals to the planes are 3, 4, – 5 and 1, 3, 3. Now, (3)(1) + (4)(3) + (– 5)(3) = 3 + 12 – 15 = 15 – 15 = 0. Therefore, the planes are perpendicular to each other. Q3. Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, ...

Q1. If f(x) = | x | and g(x) = | 3x – 4 |, find fog(x) and gof(x). Solution fog(x) = f(g(x)) = f(| 3x – 4 |) = | | 3x – 4 | | = | 3x – 4 | and gof(x) = g(f(x)) = g(|x|) = | 3| x | – 4 |. Q2. For the binary operation * defined by a * b = a + b + 1. Find the identity element for operation *. Solution Let e be the identity element for operation * and we know that a * e = a = e * a.   or a + e + 1 = a   and  e + a +1 = a or e = – 1 and e = – 1. Thus, – 1 is the identity element for operation *. Q3. If P = {5, 6} and Q = {1, 6, 7}, then find P x Q and Q x P. Solution P x Q = {(5, 1), (5, 6), (5, 7), (6, 1), (6, 6), (6, 7)} and Q x P = {(1, 5), (1, 6), (6, 5), (6, 6), (7, 5), (7, 6)} Q4. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min (a, b). Write the multiplication table for binary operation *. Also find (2 * 3) * (4 * 5). Solution  (a) We have a * b = min (a, b) on the set {1, 2, 3, 4, 5}. ...